CodeForces 414B Mashmokh and ACM （DP）

最新推荐文章于 2020-08-15 15:28:31 发布

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本文介绍了一个编程问题，任务是计算所有长度为k且满足特定整除条件的整数序列的数量。通过动态规划的方法解决了这个问题，并提供了优化的实现方案。

Mashmokh's boss, Bimokh, didn't like Mashmokh. So he fired him. Mashmokh decided to go to university and participate in ACM instead of finding a new job. He wants to become a member of Bamokh's team. In order to join he was given some programming tasks and one week to solve them. Mashmokh is not a very experienced programmer. Actually he is not a programmer at all. So he wasn't able to solve them. That's why he asked you to help him with these tasks. One of these tasks is the following.

A sequence of l integers b1, b2, ..., bl (1 ≤ b1 ≤ b2 ≤ ... ≤ bl ≤ n) is called good if each number divides (without a remainder) by the next number in the sequence. More formally $\text{[math]}$ for all i (1 ≤ i ≤ l - 1).

Given n and k find the number of good sequences of length k. As the answer can be rather large print it modulo 1000000007 (109 + 7).

Input

The first line of input contains two space-separated integers n, k (1 ≤ n, k ≤ 2000).

Output

Output a single integer — the number of good sequences of length k modulo 1000000007 (109 + 7).

Sample test(s)

input
3 2

output
5

input
6 4

output
39

input
2 1

output
2

Note

In the first sample the good sequences are: [1, 1], [2, 2], [3, 3], [1, 2], [1, 3].

大致题意：

给出一个n和k，让你从 1 -- n 这n个数中，选出 k个（可重复）组成一个序列，使这个序列满足任意一个数都能够整除该序列中它前面的那个数。求这样的序列的个数。

分析：

由于每一个数都与其前面的一个数直接相关，每个数必须是其前面一个数的倍数。那么我们可以定义一个跟末尾数字有关的状态，我们用dp[ i ] [ j ] 表示长度为 i 并且最后一位是 j 的序列，那么状态转移方程：dp [ i ] [ j ] = sum（dp [ i-1 ] [ x ] ） x 是 j 前面满足（ j % x == 0）的数。

因为是多组测试，在开始的时候直接打表可以节省时间。

比较雷的是，刚开始写的循环是

for(int i=1;i<2010;i++)

{
	for(int j=1;j<2010;j++)
	{
		for(int q=1;q<=j;q++)
		{
			if(j%q==0)
			{
				dp[i][j] += dp[i-1][q];
				dp[i][j] %= MOD;
			}
		}
	}
}

显然会T的，不能这么暴力那么无知的做。。

换成这样会好一点。至少就不会T了。。对于每一个 i ，都有从1开始的 j 去控制的 q ，来增加dp [ i ] [ ] ，这样一圈循环下来，可以完全更新~

for(int i=1;i<2010;i++)
{
	for(int j=1;j<2010;j++)
	{
		for(int q=j;q<2010;q=q+j)
		{
			dp[i][q] += dp[i-1][j];
			dp[i][q] %= MOD; 
		}
	}
}

#include 
#include 
#define MOD 1000000007

int dp[2010][2010];

int main()
{
	int n,k,sum;
	for(int i=1;i<2010;i++)
    {
        dp[1][i]=1;
    }
	for(int i=1;i<2010;i++)
	{
		for(int j=1;j<2010;j++)
		{
			for(int q=j;q<2010;q=q+j)
			{
				dp[i][q] += dp[i-1][j];
				dp[i][q] %= MOD; 
			}
		}
	}
	while(scanf("%d%d",&n,&k)!=EOF)
	{
		sum=0;
		for(int i=1;i<=n;i++)
		{
			sum=sum+dp[k][i];
			sum %= MOD;
		}
		printf("%d\n",sum);
	}
}